Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <algorithm>
#include <cmath>
#include <climits>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <cassert>
#include <functional>

using namespace std;

#define LOG(...) printf(__VA_ARGS__)
//#define LOG(...)
#define FOR(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define RSORT(c) sort((c).rbegin(),(c).rend())
#define CLR(a) memset((a), 0 ,sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int dx[] = { -1, 0, 1, 0 }; const int dy[] = { 0, 1, 0, -1 };

struct UnionFind {
	vector<int> v;
	UnionFind(int n) : v(n)  { for (int i = 0; i < n; i++) v[i] = i; }
	int  find(int x)         { return v[x] == x ? x : v[x] = find(v[x]); }
	void unite(int x, int y) { v[find(x)] = find(y); }
};

int main() {
	int n, m, start, goal;
	cin>> n>> m>> start>> goal;
	start--;
	goal--;
	vvi root(n);
	REP(i,m) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		root[a].push_back(b);
		root[b].push_back(a);
	}

	vi start_dis(n,1e9);
	vi sdn(100001,0);
	start_dis[start] = 0;
	sdn[0]++;
	queue<int> Q;
	Q.push(start);
	while (!Q.empty()) {
		int q = Q.front();Q.pop();
		REP(i, root[q].size()) {
			if (start_dis[root[q][i]] > start_dis[q] + 1) {
				start_dis[root[q][i]] = start_dis[q] + 1;
				Q.push(root[q][i]);
				sdn[start_dis[q] + 1]++;
			}
		}
	}
	vi goal_dis(n, 1e9);
	vi gdn(100001,0);
	goal_dis[goal] = 0;
	gdn[0]++;
	Q.push(goal);
	while (!Q.empty()) {
		int q = Q.front();Q.pop();
		REP(i, root[q].size()) {
			if (goal_dis[root[q][i]] > goal_dis[q] + 1) {
				goal_dis[root[q][i]] = goal_dis[q] + 1;
				Q.push(root[q][i]);
				gdn[goal_dis[q] + 1]++;
			}
		}
	}
	int dis = start_dis[goal];
	long long ercnt = 0;
	REP(i, n)
		if (start_dis[i] + goal_dis[i]+1 == dis-1)
			ercnt++;
	long long ans = 0;
	REP(i, dis-1) {
		ans += (long long)sdn[i] * (long long)gdn[dis - i-2];
	}
	cout << ans - ercnt << endl;
}